An approximation of combinatorials using the geometric mean? I'm reading a paper that uses a surprisingly accurate way of approximating a certain combinatorial structure. I'd like to try and understand how this is so effective before utilising it myself but I just can't see how it works so well!
The problem is like so:
Consider a sequence of $m$ numbers $\{ a_1, a_2, ... , a_m \}$, where $0 <a_i < 1$ for every $i$. Consider also the following expression that averages all possible  product combinations of the $a_i$ values, from $1$ to $m$. i.e.
\begin{equation}
1 + \frac{1}{\binom{m}{1}} ( a_1 + a_2 + ... + a_m ) + \frac{1}{\binom{m}{2}} ( a_1 a_2 + a_1 a_3 + ... a_{m-1} a_m ) + ... + (a_1 a_2 ... a_m)
\end{equation}
This can become incredibly computationally expensive for large $m$, so the paper says that this can be approximated using:
\begin{equation}
(a_1 a_2 ... a_m)^{0} + (a_1 a_2 ... a_m)^{\frac{1}{m}} + (a_1 a_2 ... a_m)^{\frac{2}{m}} + ... + + (a_1 a_2 ... a_m)^{1}
\end{equation}
The paper claims that this approximation is always correct to within 3% for $m$ up to 50 when tested. Anyone think they can offer an explanation as to why this is a sensible approximation? The appearance of binomial coefficients makes me think there's some sort of clever expansion being used but I can't find anything relevant.
The paper simply quotes that "A good approximation of <the above> uses the fact that the calculation of <the above> involves multiplications of all combinations of $a_i$. Therefore, we might reduce <the above> to a geometric series using the geometric average of the respective $a_i$ values."
An example for $m = 3$ is that:
\begin{equation}
1 + \frac{1}{3} (a_1 + a_2 + a_3) + \frac{1}{3} (a_1 a_2 + a_1 a_3 + a_2 a_3) + (a_1 a_2 a_3)
\end{equation}
is approximated by
\begin{equation}
(a_1 a_2 a_3)^0 + (a_1 a_2 a_3)^{\frac{1}{3}} + (a_1 a_2 a_3)^{\frac{2}{3}} + (a_1 a_2 a_3)
\end{equation}
As a final note, the application is that a competition matrix is being transformed into a transition matrix, though I've reduced the problem above to it's bare bones. Paper is here however.
Many thanks for any insight that can be given!
 A: Testing the claim for $m=2$, let $f,g$ be given by
\begin{align*}
f&=1+\Bigl({\small{\frac{1}{2}}}\Bigr)(a_1+a_2)+a_1a_2\\[4pt]
g&=1+(a_1a_2)^{\large{\frac{1}{2}}}+a_1a_2\\[4pt]
\end{align*}
where $0 < a_1,a_2 < 1$.

The author's claim is that $g$ approximates $f$ to within $3$%.

But if $a_1$ approaches $1$ from below, and $a_2$ approaches $0$ from above, the ratio ${\large{\frac{g}{f}}}$ approaches ${\large{\frac{2}{3}}}$, which is an error of approximately $33$%.

But how good would the claimed approximation be on average if $a_1,...,a_m$ are chosen independently and uniformly at random from $(0,1)$?

For each positive integer $m$, let $f_m$ be the exact expression, and let $g_m$ be the expression the author wants to use as an approximation.

For $2\le m\le 5$, using random values for $a_1,...,a_m$, we get the following simulation results . . .
\begin{array}{c|c|c}
m&\;\text{avg of}\;{\large{\frac{g_m}{f_m}}}\;{\vphantom{\frac{.}{{\LARGE{A_B}}}}}&\;\text{avg error}\;\\
\hline
2&.9632\;&3.7\text{%}\\
\hline
3&.9348&6.5\text{%}\\
\hline
4&.9108&8.9\text{%}\\
\hline
5&.8935&10.7\text{%}\\
\hline
\end{array}
${\vphantom{{\Large{\binom{A}{B}}}}}$so the author's $3$% claim appears to fail even for random values of $a_1,...,a_m$.
